Question

A 165-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s? (State the magnitude of the force.) N

Answer 1

233.3N

Step-by-step explanation:

Given:

radius 'r'= 1.5m

mass 'm'= 165kg

time 't'=2 sec

angular speed 'ω'= 0.6 rev/s

the magnitude of a torque is given by:

τ = F . r = I . α

where, 'α' is the angular acceleration and 'I' is the rotational inertia

F= I . α/ r =>[ (
`(1)/(2)` . m .
`r^(2)`)(2π . ω/t) ]/r

F=[(
`(1)/(2)` . 165 .
`1.5^(2)`)(2π . 0.6/2)]/1.5

F= 233.3N